Answer :
To determine how many grams of carbon dioxide (CO₂) are produced by burning [tex]\(65.7 \text{ g}\)[/tex] of ethane (C₂H₆) with excess oxygen, we need to follow these steps:
1. Calculate the molar mass of C₂H₆:
- Ethane (C₂H₆) consists of 2 carbon atoms and 6 hydrogen atoms.
- The atomic mass of carbon (C) is approximately 12 g/mol.
- The atomic mass of hydrogen (H) is approximately 1 g/mol.
- Therefore, the molar mass of C₂H₆ is:
[tex]\[ 2 \times 12 + 6 \times 1 = 24 + 6 = 30 \text{ g/mol} \][/tex]
2. Calculate the number of moles of C₂H₆ burned:
- The given mass of C₂H₆ is [tex]\(65.7 \text{ g}\)[/tex].
- The number of moles of C₂H₆ is calculated by dividing the mass by the molar mass:
[tex]\[ \text{moles of C₂H₆} = \frac{65.7 \text{ g}}{30 \text{ g/mol}} = 2.19 \text{ moles} \][/tex]
3. Use the stoichiometry of the reaction to find the moles of CO₂ produced:
- According to the balanced chemical equation,
[tex]\[ 2 \text{ C₂H₆} + 7 \text{ O₂} \rightarrow 4 \text{ CO₂} + 6 \text{ H₂O} \][/tex]
- This means that 2 moles of C₂H₆ produce 4 moles of CO₂.
- Therefore, the ratio of moles of CO₂ to moles of C₂H₆ is:
[tex]\[ \text{ratio} = \frac{4 \text{ moles CO₂}}{2 \text{ moles C₂H₆}} = 2 \][/tex]
- Consequently, the moles of CO₂ produced can be calculated by multiplying the moles of C₂H₆ by this ratio:
[tex]\[ \text{moles of CO₂} = 2.19 \text{ moles of C₂H₆} \times 2 = 4.38 \text{ moles} \][/tex]
4. Calculate the mass of CO₂ produced:
- The molar mass of CO₂ is:
[tex]\[ 1 \text{ carbon} = 12 \text{ g/mol} + 2 \times 16 \text{ g/mol} = 12 + 32 = 44 \text{ g/mol} \][/tex]
- The mass of CO₂ produced is calculated by multiplying the number of moles of CO₂ by its molar mass:
[tex]\[ \text{mass of CO₂} = 4.38 \text{ moles} \times 44 \text{ g/mol} = 192.72 \text{ g} \][/tex]
5. Choose the closest answer from the options provided:
The closest answer to [tex]\(192.72 \text{ g}\)[/tex] from the given options is:
[tex]\[ \boxed{192 \text{ g}} \][/tex]
Therefore, the number of grams of carbon dioxide produced by burning [tex]\(65.7 \text{ g}\)[/tex] of C₂H₆ with excess oxygen is [tex]\(\boxed{192 \text{ g}}\)[/tex].
1. Calculate the molar mass of C₂H₆:
- Ethane (C₂H₆) consists of 2 carbon atoms and 6 hydrogen atoms.
- The atomic mass of carbon (C) is approximately 12 g/mol.
- The atomic mass of hydrogen (H) is approximately 1 g/mol.
- Therefore, the molar mass of C₂H₆ is:
[tex]\[ 2 \times 12 + 6 \times 1 = 24 + 6 = 30 \text{ g/mol} \][/tex]
2. Calculate the number of moles of C₂H₆ burned:
- The given mass of C₂H₆ is [tex]\(65.7 \text{ g}\)[/tex].
- The number of moles of C₂H₆ is calculated by dividing the mass by the molar mass:
[tex]\[ \text{moles of C₂H₆} = \frac{65.7 \text{ g}}{30 \text{ g/mol}} = 2.19 \text{ moles} \][/tex]
3. Use the stoichiometry of the reaction to find the moles of CO₂ produced:
- According to the balanced chemical equation,
[tex]\[ 2 \text{ C₂H₆} + 7 \text{ O₂} \rightarrow 4 \text{ CO₂} + 6 \text{ H₂O} \][/tex]
- This means that 2 moles of C₂H₆ produce 4 moles of CO₂.
- Therefore, the ratio of moles of CO₂ to moles of C₂H₆ is:
[tex]\[ \text{ratio} = \frac{4 \text{ moles CO₂}}{2 \text{ moles C₂H₆}} = 2 \][/tex]
- Consequently, the moles of CO₂ produced can be calculated by multiplying the moles of C₂H₆ by this ratio:
[tex]\[ \text{moles of CO₂} = 2.19 \text{ moles of C₂H₆} \times 2 = 4.38 \text{ moles} \][/tex]
4. Calculate the mass of CO₂ produced:
- The molar mass of CO₂ is:
[tex]\[ 1 \text{ carbon} = 12 \text{ g/mol} + 2 \times 16 \text{ g/mol} = 12 + 32 = 44 \text{ g/mol} \][/tex]
- The mass of CO₂ produced is calculated by multiplying the number of moles of CO₂ by its molar mass:
[tex]\[ \text{mass of CO₂} = 4.38 \text{ moles} \times 44 \text{ g/mol} = 192.72 \text{ g} \][/tex]
5. Choose the closest answer from the options provided:
The closest answer to [tex]\(192.72 \text{ g}\)[/tex] from the given options is:
[tex]\[ \boxed{192 \text{ g}} \][/tex]
Therefore, the number of grams of carbon dioxide produced by burning [tex]\(65.7 \text{ g}\)[/tex] of C₂H₆ with excess oxygen is [tex]\(\boxed{192 \text{ g}}\)[/tex].